Question

Which among the following is inverse of the matrix A=\(\begin{bmatrix}2&3\\5&1\end{bmatrix}\) ?

(a) \(\begin{bmatrix}\frac{1}{13}&\frac{3}{13}\\ \frac{5}{13}&\frac{-2}{13}\end{bmatrix}\)

(b) \(\begin{bmatrix}\frac{-1}{13}&\frac{3}{13}\\ \frac{5}{13}&\frac{-2}{13}\end{bmatrix}\)

(c) \(\begin{bmatrix}\frac{-1}{13}&\frac{3}{13}\\1&\frac{-2}{13}\end{bmatrix}\)

(d) \(\begin{bmatrix}\frac{-1}{13}&\frac{3}{13}\\ \frac{5}{13}&-2\end{bmatrix}\)

I had been asked this question in examination.

I need to ask this question from Invertible Matrices topic in section Matrices of Mathematics – Class 12

Answer 1

The correct option is (b) \(\begin{bmatrix}\frac{-1}{13}&\frac{3}{13}\\ \frac{5}{13}&\frac{-2}{13}\end{bmatrix}\)

The explanation: Consider the matrix A=\(\begin{bmatrix}2&3\\5&1\end{bmatrix}\)

Using elementary row operation, we write A=IA.

\(\begin{bmatrix}2&3\\5&1\end{bmatrix}\)=\(\begin{bmatrix}1&0\\0&1\end{bmatrix}\)A

\(\begin{bmatrix}-13&0\\5&1\end{bmatrix}\)=\(\begin{bmatrix}1&-3\\0&1\end{bmatrix}\)A   (Applying R1→R1-3R2)

\(\begin{bmatrix}1&0\\5&1\end{bmatrix}\)=\(\begin{bmatrix}\frac{-1}{13}&\frac{3}{13}\\0&1\end{bmatrix}\)A   (Applying \(R_1 \rightarrow -\frac{R_1}{13}\))

\(\begin{bmatrix}1&0\\0&1\end{bmatrix}\)=\(\begin{bmatrix}\frac{-1}{13}&\frac{3}{13}\\ \frac{5}{13}&\frac{-2}{13}\end{bmatrix}\)A   (Applying R2→R2-5R1)

\(A^{-1}=\begin{bmatrix}\frac{-1}{13}&\frac{3}{13}\\ \frac{5}{13}&\frac{-2}{13}\end{bmatrix}\).